$\text{ If }{S}_n\text{ denotes the sum of first }n\text{ terms of an A.P whose first term is }a\text{ and }\frac{S_{nx}}{S_x}\text{ is independent of }\mathrm{x}\text{, then }{S}_{\rho }=$

$\begin{array}{l}\begin{array}{ll}\frac{S_{nx}}{S_x}=\frac{\frac{nx}{2}[2a+(nx-1\left)d\right]}{\frac{x}{2}[2a+(x-1\left)d\right]}& \\ & =\frac{n\left[\right(2a-d)+nxd]}{(2a-d)+xd}\end{array}\\ \text{ For }\frac{S_{nx}}{S_x}\text{ to be independent of }\mathrm{x}\text{, }\end{array}$

$\begin{array}{l}2a-d=0\Rightarrow 2a=d\\ \text{ Now, }\\ S_p=\frac{p}{2}[2a+(p-1\left)d\right]=p^{2}a\end{array}$